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  • 数据库相同成绩排名例子

    数据库相同成绩排名例子

    create table score1(
           s_id number,
           s_score decimal(5,2) --表示最大5位的数字,其中小数点后面两位小数
    )
    insert into score1 (s_id, s_score) values ('1', '3.5');
    insert into  score1 (s_id, s_score) values ('2', '3.65');
    insert into  score1 (s_id, s_score) values ('3', '4.0');
    insert into score1 (s_id, s_score) values ('4', '3.85');
    insert into  score1 (s_id, s_score) values ('5', '4.0');
    insert into  score1 (s_id, s_score) values ('6', '3.65');
    select *from score1
    select s_score,(select count(distinct s_score)from score1 where s_score>=s.s_score)排名 from score1 s order by s_score desc
    select count(distinct s_score)from score1

     查询表中排名前三的数据

    Create table  Employee (Id int, Name varchar(255), Salary int, DepartmentId int);
    Create table  Department (Id int, Name varchar(255));
    
    select *from employee
    insert into Employee (Id, Name, Salary, DepartmentId) values ('1', 'Joe', '70000', '1');
    insert into Employee (Id, Name, Salary, DepartmentId) values ('2', 'Henry', '80000', '2');
    insert into Employee (Id, Name, Salary, DepartmentId) values ('3', 'Sam', '60000', '2');
    insert into Employee (Id, Name, Salary, DepartmentId) values ('4', 'Max', '90000', '1');
    insert into Employee (Id, Name, Salary, DepartmentId) values ('5', 'Janet', '69000', '1');
    insert into Employee (Id, Name, Salary, DepartmentId) values ('6', 'Randy', '85000', '1');
    
    insert into Department (Id, Name) values ('1', 'IT');
    insert into Department (Id, Name) values ('2', 'Sales');
    select *from Department
    
    select d.Name as Department,e1.Name as Employee,e1.Salary 
    from Employee e1 join
     Department d
    on d.Id=e1.DepartmentId
     where 3>
    (select count(distinct e2.Salary)from Employee e2 where e2.Salary>e1.Salary 
    and e1.departmentid=e2.departmentid)
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  • 原文地址:https://www.cnblogs.com/tangjiang-code/p/7668393.html
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